Worked solution
Condition on the first black draw before counting the remaining whites
A bag initially contains 3 white and 6 black balls. A player draws 3 balls, one at a time, without replacement. Three white balls win the grand prize; exactly two whites win a consolation prize; otherwise the player wins nothing.
Find the probability that
(i) a player wins the grand prize,
(ii) a player wins a consolation prize,
(iii) a player wins a consolation prize, given that the first draw is black.
CJC · Review problems · Question 11
Answer
(i) ; (ii) ; (iii) .
Name the ordered-draw events
Let Wᵢ and Bᵢ denote a white or black ball on draw i. Counts and the total decrease after every draw because there is no replacement.
(i) All three whites
.
(ii) Exactly two whites in any order
The source counts W₁W₂B₃ then its distinct colour orders. Their products are equal despite differently ordered intermediate counts. Hence .
(iii) Condition on B₁
The joint event is B₁W₂W₃. Thus
The first black ball leaves all three whites among eight balls.
Common mistakes
Do not multiply by three colour orders after conditioning on black first: only BWW remains.
Solution notes
All three asks, source event definitions, every ordered product/multiplicity and full conditional cancellation retained. No marks printed.